设f(x)满足方程af(x)+bf(x/x-1)=e^x,其中|a|不等于|b|,求f(x)

一看这种类型的题目，如af(x)+bf(x/x-1)=e^x
就直接把x作为x/x-1。
代入x/x-1，则方程变为 af(x/x-1)+bf(x)=e^(x/x-1)
然后，将f(x/x-1)用f(x)与x表达出来就有：
f(x/x-1)=e^(x/x-1) /a - bf(x)/a
代入af(x)+bf(x/x-1)=e^x 则有：
af(x)+(b/a)e^(x/x-1)-(b²/a)f(x)=e^x
(a²-b²)/a f(x)=e^x-(b/a)e^(x/x-1)
f(x)={ae^x-be^(x/x-1)}/(a²-b²)

设g(x)=x/(x-1),则
g[g(x)]=x.
af(x)+bf[x/(x-1)]=e^x,①
af[x/(x-1)]+bf(x)=e^[x/(x-1)],②
①*a-②*b,得
(a^2-b^2)f(x)=ae^x-be^[x/(x-1)],
因|a|≠|b|,故
f(x)={ae^x-be^[x/(x-1)]}/(a^2-b^2).

在af(x)+bf(x/(x-1))=e^x中,x≠1,且x/(x-1) ≠1设t=x/(x-1),得x=t/(t-1),∴af(t)+bf(t/(t-1))=e^t ①af(t/(t-1))+bf(t)=e^[ t/(t-1)] ②①×a-②×b,得(a²-b²)f(t)=a×e^t-b×e^[ t/(t-1)]∵|a|≠|b|,∴a&...

令y=x/(x-1),得x=y/(y-1)
代入原方程，得
af[y/(y-1)]+bf(y)=e^[y/(y-1)]
将参数y换成x，得
af[x/(x-1)]+bf(x)=e^[x/(x-1)]
与原方程联立，得
f(x)={ae^x-be^[x/(x-1)]}/(a^2-b^2)